Leakage from underground gasoline tanks at service stations can damage the environment. It is estimated that 25% of these tanks leak. You examine 15 tanks chosen at random, independently of each other.

(a). What is the mean number of leaking tanks in such samples of 15?

(b). What is the probability that 10 or more of the tanks leak?

(c). Now you do a larger study, examining a random sample of 1000 tanks nationally. What is the probability that at least 275 of these tanks are leaking?

a) Mean of leaked tanks in a sample size of 15 = 3.75

b) Probability that 10 or more of the tanks out of 15 will leak = 0.000795

c) 0.034

Step-by-step explanation:

a) Mean number of leaking tanks in a sample size = np

n = sample size

p = probability of a leagage

Mean = 0.25× 15 = 3.75

b) The probability that 10 or more of the tanks leak

This is a binomial distribution problem

Binomial distribution function is represented by

P(X = x) = ⁿCₓ pˣ qⁿ⁻ˣ

n = total number of sample spaces = 15 tanks

x = Number of successes required = number of leaked tanks expected

p = probability of success = probability of a tank leaking = 0.25

q = probability of failure = 1 - p = 0.75

P(X ≥ 10) = P(X=10) + P(X=11) + P(X=12) + P(X=13) + P(X=14) + P(X=15)

P(X ≥ 10) = 0.000795

c) For a sample size of 1000, we first investigate if it follows a normal distribution as it satisfies the two required conditions for a normal distribution

np = 0.25 × 1000 = 250 > 10

np(1-p) = 1000 × 0.25 × 0.75 = 187.5 > 10

Mean = np = 250

Standard deviation = √[np(1-p)] = √(187.5) = 13.70

The probability that at least 275 of these tanks are leaking = P(x ≥ 275)

We first standardize 275

The standardized score for a value is the value minus the mean then divided by the standard deviation.

z = (x - μ)/σ = (275 - 250)/13.7 = 1.825

P(x ≥ 275) = P(z ≥ 1.825)

We'll use data from the normal probability table for these probabilities

P(x ≥ 275) = P(z ≥ 1.825) = P(z ≤ - 1.825) = 0.034

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step-by-step explanation: